I don't know guys.

I have everything hooked up like in the 1st diagram (minus the transistor and relay) and the output (pin 3) of the 2nd 555 goes to an LED. The light just stays on and is very dim.

About an hour ago I was able to tap +9V to the trigger of the 2nd 555 and the light would turn on and flash, but it never truly turned off, it just went dim and bright back and forth. The trigger for the 1st 555 never did anything. Now, touching +9V to the 2nd 555 turns it off. I wonder if both the 555s are fried. I verified the LED was OK by going directly from the battery to it and it's really bright, so why it just stays on now is a mystery to me. The black caps on the right are 0.47uf.

Anther angle

 

I isolated the 555s and just ran 9V to the power (pin 8) and the reset (pin 4) and checked the output of pin 3. It's 9V, but the light is very dim like I mentioned in the post above. It's doing that on both 555s. How is that possible? Is it possible a resistor or capacitor shorted when the wires touched?
Pic #1

I then jumped the LED directly to the 9V and it's working normally.
Pic #2

 

Your understanding of the 555 inputs needs some work. Pin 2 is an input, and it can swing from GND to Vcc. Also, it is the trigger input; the circuit will not function without it being driven by a signal. Often it is pulled up to Vcc as the rest state, and pulled down to GND to trigger the monostable. The diode has no role in this. The diode is there to protect the pin 2 input from voltage transients that exceed Vcc, which are common in automotive power systems. Because it is reverse biased as drawn, it is doing everything to prevent +12 from coming into the Trigger input.

If the trigger input spikes above 12.6 V, the diode conducts and clamps the input at approximately this voltage. This prevents interdamage in the 555.

ak

 

Just real quick to respond to this, the diagram shows two paths to allow 12V into the pin, so the diode is useless. The resistor and diode are touching the same row on the breadboard in that diagram.

The diode is there to protect the pin 2 input from voltage transients that exceed Vcc, which are common in automotive power systems. Because it is reverse biased as drawn, it is doing everything to prevent +12 from coming into the Trigger input.

If the trigger input spikes above 12.6 V, the diode conducts and clamps the input at approximately this voltage. This prevents interdamage in the 555.

ak

 

Just real quick to respond to this, the diagram shows two paths to allow 12V into the pin, so the diode is useless. The resistor and diode are touching the same row on the breadboard in that diagram.

AnalogKid is working hard to give you solid advice here. You should try being a little more patient and read through what he said. He already answered this question. Yes, there is a path for 12V to get to that pin and no, that doesn't make the diode useless. The diode isn't there to prevent 12V from reaching that pin. It's there so that if the voltage on that pin is pulled significantly ABOVE 12V, the diode will carry that excess voltage away, protecting the input.

It's perfectly natural for you to be confused if clamping diodes are new to you, but don't lash out at people who are helping you, certainly not by correcting them on aspects of electronics that they clearly have more experience with than you do.

With just a little patience and respect you'll find that this forum will be incredibly helpful.

 

why is there a diode AND a resistor on the +12V rail?

The diode blocks any negative-going load-dump voltage spikes on the positive rail. The 10Ω resistor limits current which the TVS might have to pass if there is a positive-going spike. The 100Ω resistor and 220uF cap smooth out the supply voltage for the rest of the circuit.

 

Not quite. The diode clamps positive going inductive spikes.

ak

 

ebeowolf: I see your point, things get lost in translation in text. I am frustrated, especially now that the project isn't working at all. Both 555s are not responding. Apologies to AK if it appears I'm lashing out, but I don't feel that way as an actual person. I started this 2 weeks ago and have been struggling to understand the process. If I knew anything about electronics, this would have been done in about 45 minutes, but now I'm on 2 weeks and almost $80 in parts. At this level, I understand a diode allows electricity to flow in one direction, so at that point I see a diode and an open line to a pin inline with a diode, I have to question it because that's all my limited knowledge understands. You see where I'm coming from? Part of the problem is that I have zero fundamental understanding of these concepts and have difficulty learning on my own, I need someone in person visually explaining every piece and reading leaves me up to my own interpretations, which is not advised.

Would anyone consider allowing me to ship the board to them? I don't want someone else to build it, but I want them to look at it and tell me what's wrong. I don't know of any local electronic clubs where I am and showing this to someone without an experienced EE background may backfire, they might take it apart and do it wrong. Or, is it possible for someone to make a simplified drawing of this so I can start over without the diodes and resistors just so I can see how it's supposed to work on 9V?

I've provided pictures and can continue to provide more if needed. Should I just go out and buy 2 more 555s? I'm afraid if do that, the current wiring is wrong and they'll blow up again. I don't even know for sure if they are blown.

 

ebeowolf: I see your point, things get lost in translation in text. I am frustrated, especially now that the project isn't working at all. Both 555s are not responding. Apologies to AK if it appears I'm lashing out, but I don't feel that way as an actual person. I started this 2 weeks ago and have been struggling to understand the process. If I knew anything about electronics, this would have been done in about 45 minutes, but now I'm on 2 weeks and almost $80 in parts. At this level, I understand a diode allows electricity to flow in one direction, so at that point I see a diode and an open line to a pin inline with a diode, I have to question it because that's all my limited knowledge understands. You see where I'm coming from? Part of the problem is that I have zero fundamental understanding of these concepts and have difficulty learning on my own, I need someone in person visually explaining every piece and reading leaves me up to my own interpretations, which is not advised.

Would anyone consider allowing me to ship the board to them? I don't want someone else to build it, but I want them to look at it and tell me what's wrong. I don't know of any local electronic clubs where I am and showing this to someone without an experienced EE background may backfire, they might take it apart and do it wrong. Or, is it possible for someone to make a simplified drawing of this so I can start over without the diodes and resistors just so I can see how it's supposed to work on 9V?

I've provided pictures and can continue to provide more if needed. Should I just go out and buy 2 more 555s? I'm afraid if do that, the current wiring is wrong and they'll blow up again. I don't even know for sure if they are blown.

No worries! I understand how frustrating these things can be. I have a hard time reading breadboard circuits like in your pics, but I'm going to give it my best shot. Hopefully it's something simple, like swapped pins or a short somewhere. I'll let you know right away if I find anything.

 

checked the output of pin 3. It's 9V, but the light is very dim

It depends what your "LED" is. From the pic it looks like a set of LEDs lit simultaneously. If the LEDs are white ones in series it's not surprising that they are dim (each drops ~3V when lit). A 555 pin 3 will probably drop to about 7.8V under load. Presumably the set has an inbuilt current-limiting resistor, since you haven't used one between the set and pin 3 from what I can see?
What I suggest you do is start over, clear the breadboard and get just the first 555 doing its thing. Once you know that's working you can add the second 555. Building circuits can be frustrating. If you break the project down into parts it becomes more manageable. Don't forget the 555 needs pin 2 voltage to be pulled down to near ground (the negative rail) to trigger the 555 pulse. If you use a push-button for that, on the first 555, connect it from the left end of C1 to ground, and connect a 10k (-ish) resistor from that left end to the +9V supply.

I don't know of any local electronic clubs where I am

And we don't know where you are .

 

It's hard to tell for sure in those pics, but it looks like I see a short jumper wire joining pins 6&7 (threshold and discharge) on each of the two 555 chips. There should be a 1k resistor between those pins, not a jumper. I apologize if I'm just seeing the images wrong. I'd double check those connections (and related connections - R2, R3, C3, R5, R6, C6) first.

The suggestion above to work on one 555 at a time seems like a good idea - when a project gets troublesome like this, it's best to break it into smaller pieces in order to troubleshoot simpler systems independently.

Finally, this last thought is really a question for the more experienced members of this forum, cause I still have a LOT to learn! Shouldn't there be individual decoupling caps for each 555? I know the requirements for decoupling vary somewhat depending on the application, but I expected to see one low value (anywhere from 0.01uF to 1uF?) cap for each 555, as close as possible to its VCC input (pin 8.) Is that not necessary in this situation?

 

There is a jumper wire there. I did that based on this (below), which shows pin 7 going up into pin 6, so I put a jumper. There are also 2 resistors on that same horizontal line as pin 7 which doesn't make sense to me, why have two resistors on one line, why not use 1 with the correct capacity? Again, I don't know what I'm doing.

 

There is a jumper wire there. I did that based on this (below), which shows pin 7 going up into pin 6, so I put a jumper. There are also 2 diodes on that same horizontal line as pin 7 which doesn't make sense to me, why have two resistors on one line, why not use 1 with the correct capacity? Again, I don't know what I'm doing.

Oops, sorry! I was checking it against Alec's second schematic, not the first. I wonder which one it's supposed to be. Hopefully Alec will clarify.

 

Ok, I'm looking carefully at the pics now and comparing to the first schematic this time. I see a few apparent discrepancies:
1) On the discharge (pin 7) of each 555, I see one resistor with both legs in the same row, which effectively short circuits right past that resistor, and then there are the expected capacitor and resistor beyond. Taking the first 555 as an example, what should be happening there is one leg of the resistor R2 should be in the row with pin 7, and the other leg of that resistor should connect to a new, previously unoccupied row. The connections to R3 and C3 should be made from that row.
2) There appears to be a straight wire from U1 out to U2 trigger. There is supposed to be a capacitor C4 in series there. This means a similar change as described above. Run a wire from the row with U2 output pin to a new, unoccupied row. Then connect the capacitor C4 from that new row to U2 trigger row, where connections with R4 and D2 are also made.

I may be missing something else, but that's all my sleepy eyes can find for now. Best of luck. Let me know if those changes improve things at all. If not, maybe upload new pics reflecting the changes and I'll take another, closer look.

 

There is a jumper wire there. I did that based on this (below), which shows pin 7 going up into pin 6, so I put a jumper. There are also 2 resistors on that same horizontal line as pin 7 which doesn't make sense to me, why have two resistors on one line, why not use 1 with the correct capacity? Again, I don't know what I'm doing.

As for the two resistors in series question, that can legitimately happen for more than one reason. One situation is when a single resistor is not available in the exact value required (or is rare, expensive, etc) but can be replaced with two common values that add up to the desired value.

In this case, it's because part of this circuit requires a connection to the junction of those two resistors (C3 connects at the intersection of R2 and R3.) If R2 and R3 were replaced with a single resistor of equivalent total value, there would be no appropriate place to connect C3 for the desired result.

 

I was checking it against Alec's second schematic, not the first. I wonder which one it's supposed to be. Hopefully Alec will clarify.

My bad in the first (late-night!) schematic (though it shoudn't noticeably affect circuit operation). The second schematic is correct; pin 6 should connect to the top of the cap (C3/C6), not to pin 7.
As ebeowulf17 pointed out, 'decoupling' capacitors (which act as a small reservoir of charge to help stabilise the 555 supply) might be needed. If so, a respective 100nF cap should be connected between pins 1 and 8 of each 555 (close to the pins).

 

Having a strange experience now. I took everything apart and started over with the second diagram. When I plug in the Vcc, the light turns on. Does not blink. But if I touch the board, the light turns off. It's being grounded by me touching the board I guess?
Here is a picture.

Here is a shot where the light is on,

and another shot (now the light is off as I touched the board) up close

 

The positive rail on the left is still being controlled by the momentary switch, right? If so, that means that when the switch isn't being pressed, it's not connected to power supply, ground, or any other known voltage. In other words, it's "floating." IC inputs REALLY don't like to float. They need to always be in a known, deliberate, controlled state. The resistor from pin 2 of 555 to left positive rail just needs to move to right positive rail in order to provide a pull up to hold that pin high (except for the occasional pulse through the capacitor when the button is pressed and released.)

What's the big cap by the LED for? Presumably it's the cap that will turn the first 555's output into pulses for the second 555 later. Whatever it is, it's doing nothing as long as both of its legs are in the same row.

Otherwise I think you're really close. I'm guessing the finger control you're getting is because the floating input is really easy to bias one way or another. I think the apparent dim LED lighting is probably actually incredibly fast cycling on and off and that your finger stabilizes the input which in turn stops the intermittent output. Fingers crossed!

 

It's not clear from the pics; have you got a +ve supply rail and a ground rail actually connected on both sides of the breadboard? I would expect to see a red wire (+ve) left and right and a black wire (ground) left and right.

 

I now have the +/- on one side from the battery, and they are jus jumped over to the other side. Is that wrong?

Here is a good picture. This should be easy to see and point out exactly what's wrong. Now the light doesn't even turn on at all.